Question

Please solve question number 16 with explanation , urgent . question in the picture

Please No Spam

Answer 1

Hey

$here \: i s \: t he\ \: answer$



Distance = 1600 km

Let the usual speed be = x km/h

now we know that 

speed = distance /time

time = distance / speed

Usual time = 1600/x

Now due to bad weather the speed is increased b 400 km/ hr

which means new speed = 1600/x+400

Now it's given that the plain left 40 minutes late

which means new time = 1600/x+400 + 2/3 ( 40/60 = 2/3 hours)

1600/x = 1600/x+400 +2/3

1600/x-1600/x+400 = 2/3

1600(x+400)-1600x)/x(x+400) = 2/3

1600x + 640000-1600x/x²+400x = 2/3

640000 = 2/3(x²+400x)

640000 * 3/2 = x²+400x

960000 = x² +400x

0 = x²+400x-960000

0 = x² +1200x-800x-960000

0 = x( x+ 1200) - 800(x+1200)

0= (x+1200)(x-800)

x= -1200 or x= 800

Speed cannot be negative 
∴ usual speed = 800 km/h

Hope it helps! 

Answer 2

Distance = 1600 km

Let the usual speed be = x km/h

now we know that 

speed = distance /time

time = distance / speed

Usual time = 1600/x

Now due to bad weather the speed is increased b 400 km/ hr

which means new speed = 1600/x+400

Now it's given that the plain left 40 minutes late

which means new time = 1600/x+400 + 2/3 ( 40/60 = 2/3 hours)

1600/x = 1600/x+400 +2/3

1600/x-1600/x+400 = 2/3

1600(x+400)-1600x)/x(x+400) = 2/3

1600x + 640000-1600x/x²+400x = 2/3

640000 = 2/3(x²+400x)

640000 * 3/2 = x²+400x

960000 = x² +400x

0 = x²+400x-960000

0 = x² +1200x-800x-960000

0 = x( x+ 1200) - 800(x+1200)

0= (x+1200)(x-800)

x= -1200 or x= 800

Speed cannot be negative 
∴ usual speed = 800 km/h